Answer:
Step-by-step explanation:
[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]
Recall that we can express the squared trig functions in terms of cos(2x). That is,
And so inverting these,
.
Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).
Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).
So then,
This is now integrable (phew),
Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
Answer:
<h3>3.75</h3>
Step-by-step explanation:
Using Secant-Secant theorem we can find the value of x.
The product of one segment and its external segment is equal to the product of the other segment and its external segment.
5 × 3 = x × 4
15 = 4x
15/4 = x
3.75 = x
